Sum of minimum absolute difference of each array element

Given an array of n distinct integers. The problem is to find the sum of minimum absolute difference of each array element. For an element x present at index i in the array its minimum absolute difference is calculated as: 
Min absolute difference (x) = min(abs(x – arr[j])), where 1 <= j <= n and j != i and abs is the absolute value. 
Input Constraint: 2 <= n

Examples: 

Input : arr = {4, 1, 5}
Output : 5
Sum of absolute differences is |4-5| + |1-4| + |5-4|

Input : arr = {5, 10, 1, 4, 8, 7}
Output : 9

Input : {12, 10, 15, 22, 21, 20, 1, 8, 9}
Output : 18

Naive approach: Using two loops. Pick an element of the array using outer loop and calculate its absolute difference with rest of the array elements using inner loop. Find the minimum absolute value and add it to the sum. Time Complexity O(n²). 

C++

// C++ implementation to find the sum
// of minimum absolute difference of
// each array element
#include <bits/stdc++.h>
using namespace std;

// function to find the sum of
// minimum absolute difference
int sumOfMinAbsDifferences(int arr[], int n)
{
  int sum = 0;
  for (int i = 0; i < n; i++) {
    int diff = INT_MAX;
    for (int j = 0; j < n; j++) {
      if (i != j) {
        diff = min(diff, abs(arr[i] - arr[j]));
      }
    }
    sum += diff;
  }

  // required sum
  return sum;
}

// Driver code
int main()
{
  int arr[] = { 5, 10, 1, 4, 8, 7 };
  int n = 6;

  cout << "Sum = " << sumOfMinAbsDifferences(arr, n);
}

// This code is contributed by garg28harsh.

Java

// java implementation to find the sum
// of minimum absolute difference of
// each array element

import java.util.*;
import java.io.*;
public class GFG {

    // function to find the sum of
    // minimum absolute difference
    static int sumOfMinAbsDifferences(
                        int arr[] ,int n)
    {
        int sum=0;
      for(int i=0;i<n;i++){
        int diff=Integer.MAX_VALUE;
       for(int j=0;j<n;j++){
        if(i!=j){
          diff=Math.min(diff,Math.abs(arr[i]-arr[j]));
        }
       }
        sum+=diff;

      }
        // required sum
        return sum;
    }

    // Driver code
    public static void main(String args[])
    {
        int arr[] = {5, 10, 1, 4, 8, 7};
        int n = arr.length;

        System.out.println( "Sum = "
        + sumOfMinAbsDifferences(arr, n));
    }
}

Python3

# Python3 implementation to find the sum
# of minimum absolute difference of
# each array element
import sys

# function to find the sum of
# minimum absolute difference
def sumOfMinAbsDifferences(arr, n):
    sum = 0
    for i in range(n):
        diff = sys.maxsize
        for j in range(n):
            if i != j:
                diff = min(diff, abs(arr[i] - arr[j]))
        sum += diff

    # required sum
    return sum

# Driver code
if __name__ == "__main__":
    arr = [5, 10, 1, 4, 8, 7]
    n = 6

    print("Sum =", sumOfMinAbsDifferences(arr, n))

# This code is contributed by akashish__

C#

// Include namespace system
using System;

public class GFG
{
  // function to find the sum of
  // minimum absolute difference
  public static int sumOfMinAbsDifferences(int[] arr, int n)
  {
    var sum = 0;
    for (int i = 0; i < n; i++)
    {
      var diff = int.MaxValue;
      for (int j = 0; j < n; j++)
      {
        if (i != j)
        {
          diff = Math.Min(diff,Math.Abs(arr[i] - arr[j]));
        }
      }
      sum += diff;
    }
    // required sum
    return sum;
  }

  // Driver code
  public static void Main(String[] args)
  {
    int[] arr = {5, 10, 1, 4, 8, 7};
    var n = arr.Length;
    Console.WriteLine("Sum = " + GFG.sumOfMinAbsDifferences(arr, n).ToString());
  }
}

// This code is contributed by aadityaburujwale.

Javascript

// Javascript implementation to find the sum
// of minimum absolute difference of
// each array element

    // function to find the sum of
    // minimum absolute difference
    function sumOfMinAbsDifferences(arr, n)
{
    let sum = 0;
    for (let i = 0; i < n; i++) {
        let diff = Number.MAX_VALUE;
        for (let j = 0; j < n; j++) {
            if (i != j) {
                diff = Math.min(diff,
                                Math.abs(arr[i] - arr[j]));
            }
        }
        sum += diff;
    }

    // required sum
    return sum;
}

// Driver code
let arr = [ 5, 10, 1, 4, 8, 7 ];
let n = 6;

console.log("Sum = " + sumOfMinAbsDifferences(arr, n));

// This code is contributed by garg28harsh.

Output

Sum = 9

Time Complexity: O(n²) where n is size of the input array. This is because two nested loops are executing.

Space Complexity: O(1) as  no extra space has been used.

Efficient Approach:

The following steps are:

1. Sort the array of size n.
2. For the 1st element of array its min absolute difference is calculated using the 2nd array element.
3. For the last array element its min absolute difference is calculated using the 2nd last array element.
4. For the rest of the array elements, 1 <= i <= n-2, minimum absolute difference for an element at index i is calculated as: minAbsDiff = min( abs(arr[i] – arr[i-1]), abs(ar[i] – arr[i+1]) ).

Implementation:

C++

// C++ implementation to find the sum of minimum
// absolute difference of each array element
#include <bits/stdc++.h>
using namespace std;

// function to find the sum of
// minimum absolute difference
int sumOfMinAbsDifferences(int arr[], int n)
{
    // sort the given array
    sort(arr, arr+n);

    // initialize sum
    int sum = 0;

    // min absolute difference for
    // the 1st array element
    sum += abs(arr[0] - arr[1]);

    // min absolute difference for
    // the last array element
    sum += abs(arr[n-1] - arr[n-2]);

    // find min absolute difference for rest of the
    // array elements and add them to sum
    for (int i=1; i<n-1; i++)
        sum += min(abs(arr[i] - arr[i-1]), abs(arr[i] - arr[i+1]));

    // required sum
    return sum;
}

// Driver program to test above
int main()
{
    int arr[] = {5, 10, 1, 4, 8, 7};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Sum = "
         << sumOfMinAbsDifferences(arr, n);
}

Java

// java implementation to find the sum
// of minimum absolute difference of
// each array element
import java.io.*;
import java.util.Arrays;

public class GFG {

    // function to find the sum of
    // minimum absolute difference
    static int sumOfMinAbsDifferences(
                         int arr[] ,int n)
    {

        // sort the given array
        Arrays.sort(arr);

        // initialize sum
        int sum = 0;

        // min absolute difference for
        // the 1st array element
        sum += Math.abs(arr[0] - arr[1]);

        // min absolute difference for
        // the last array element
        sum += Math.abs(arr[n-1] - arr[n-2]);

        // find min absolute difference for
        // rest of the array elements and
        // add them to sum
        for (int i = 1; i < n - 1; i++)
            sum +=
            Math.min(Math.abs(arr[i] - arr[i-1]),
                    Math.abs(arr[i] - arr[i+1]));

        // required sum
        return sum;
    }

    // Driver code
    public static void main(String args[])
    {
        int arr[] = {5, 10, 1, 4, 8, 7};
        int n = arr.length;

        System.out.println( "Sum = "
        + sumOfMinAbsDifferences(arr, n));
    }
}

// This code is contributed by Sam007.

Python3

# Python implementation to find the
# sum of minimum absolute difference
# of each array element

# function to find the sum of
# minimum absolute difference

def sumOfMinAbsDifferences(arr,n):
    # sort the given array
    arr.sort()
    # initialize sum
    sum = 0

    # min absolute difference for
    # the 1st array element
    sum += abs(arr[0] - arr[1]);

    # min absolute difference for
    # the last array element
    sum += abs(arr[n - 1] - arr[n - 2]);

    # find min absolute difference for
    # rest of the array elements and
    # add them to sum
    for i in range(1, n - 1):
        sum += min(abs(arr[i] - arr[i - 1]),
                  abs(arr[i] - arr[i + 1]))

    # required sum
    return sum;


# Driver code
arr = [5, 10, 1, 4, 8, 7]
n = len(arr)
print( "Sum = ", sumOfMinAbsDifferences(arr, n))


#This code is contributed by Sam007

C#

// C# implementation to find the sum
// of minimum absolute difference of
// each array element
using System;

public class GFG {

    // function to find the sum of
    // minimum absolute difference
    static int sumOfMinAbsDifferences(
                         int []arr ,int n)
    {

        // sort the given array
        Array.Sort(arr);

        // initialize sum
        int sum = 0;

        // min absolute difference for
        // the 1st array element
        sum += Math.Abs(arr[0] - arr[1]);

        // min absolute difference for
        // the last array element
        sum += Math.Abs(arr[n-1] - arr[n-2]);

        // find min absolute difference for
        // rest of the array elements and
        // add them to sum
        for (int i = 1; i < n - 1; i++)
            sum +=
            Math.Min(Math.Abs(arr[i] - arr[i-1]),
                    Math.Abs(arr[i] - arr[i+1]));

        // required sum
        return sum;
    }

    // Driver code
    public static void Main ()
    {
        int []arr = {5, 10, 1, 4, 8, 7};
        int n = arr.Length;

        Console.Write( "Sum = "
        + sumOfMinAbsDifferences(arr, n));
    }
}

// This code is contributed by Sam007.

PHP

<?php
// PHP implementation to find
// the sum of minimum absolute
// difference of each array element

// function to find the sum of
// minimum absolute difference
function sumOfMinAbsDifferences($arr, $n)
{

    // sort the given array
    sort($arr);
    sort( $arr,$n);

    // initialize sum
    $sum = 0;

    // min absolute difference for
    // the 1st array element
    $sum += abs($arr[0] - $arr[1]);

    // min absolute difference for
    // the last array element
    $sum += abs($arr[$n - 1] - $arr[$n - 2]);

    // find min absolute difference
    // for rest of the array elements
    // and add them to sum
    for ($i = 1; $i < $n - 1; $i++)
        $sum += min(abs($arr[$i] - $arr[$i - 1]),
                   abs($arr[$i] - $arr[$i + 1]));

    // required sum
    return $sum;
}

    // Driver Code
    $arr = array(5, 10, 1, 4, 8, 7);
    $n = sizeof($arr);
    echo "Sum = ", sumOfMinAbsDifferences($arr, $n);

// This code is contributed by nitin mittal.
?>

Javascript

<script>

    // Javascript implementation to find the sum
    // of minimum absolute difference of
    // each array element

    // function to find the sum of
    // minimum absolute difference
    function sumOfMinAbsDifferences(arr, n)
    {

        // sort the given array
        arr.sort(function(a, b){return a - b});

        // initialize sum
        let sum = 0;

        // min absolute difference for
        // the 1st array element
        sum += Math.abs(arr[0] - arr[1]);

        // min absolute difference for
        // the last array element
        sum += Math.abs(arr[n-1] - arr[n-2]);

        // find min absolute difference for
        // rest of the array elements and
        // add them to sum
        for (let i = 1; i < n - 1; i++)
            sum +=
            Math.min(Math.abs(arr[i] - arr[i-1]),
                    Math.abs(arr[i] - arr[i+1]));

        // required sum
        return sum;
    }

    let arr = [5, 10, 1, 4, 8, 7];
    let n = arr.length;

    document.write( "Sum = " + sumOfMinAbsDifferences(arr, n));

</script>

Output

Sum = 9

Time Complexity: O(n log n)
Auxiliary Space: O(1)

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.